# Falling Body with Zero, Linear, or Quadratic Air Resistance

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This Demonstration shows plots of the one-dimensional downward trajectory of a falling body as a function of time. The cases considered are: no air resistance (blue), linear resistance (red), and quadratic resistance (purple). The drop height (vertical axis) is in meters and the horizontal axis is time in seconds. The acceleration due to gravity is set to . You can vary the drop height , the terminal velocity for linear air resistance , the terminal velocity for quadratic air resistance , and the time , in order to see where the object is at a certain time.

Contributed by: Anne Tabor-Morris (May 2014)

Georgian Court University

Open content licensed under CC BY-NC-SA

## Snapshots

## Details

A body falling under gravity in a vacuum (i.e. encountering no air resistance) falls at the maximum rate (so that the time of the fall is minimized) and never attains terminal velocity. This Demonstration plots the position versus time using the Earth's acceleration due to gravity .

However, if a falling body encounters air resistance, it slows down when compared to the fall of that same object in a vacuum. Air resistance can require a linear or quadratic correction, depending on the diameter and speed of the object. Sometimes there is a combined linear and quadratic effect (not explored here).

Let be the mass of the object; in all three cases, , where is the force and is the acceleration.

The equation of motion for no air resistance is

.

For linear air resistance ( is the velocity and is the linear term; see more details below):

.

For quadratic air resistance ( is the quadratic term):

.

To find the position of the object, the equations of motion ( is the acceleration) are integrated twice. The first integration gives the velocity of the object and the second determines the position of the object as a function of time.

This is the position with no air resistance: .

This is the position with linear resistance: , where the terminal velocity .

This is the position with quadratic resistance: , where the terminal velocity .

For a full explanation on how to solve the equations of motion, see pp. 53 and 61 of [1].

This Demonstration shows that the time it takes for an object released from rest at a given height, falling in Earth's downward gravity field with no air resistance, is less than that for an object undergoing air resistance. The time differs for linear or quadratic air resistance; the terminal velocities depend on the diameter of the object (assumed spherical), as well as on the viscosity of the medium, which, if air, is usually assumed to be at standard temperature and pressure (STP).

You can vary the drop height. You can also vary the time to show where the same object is in its flight; extend time until a curve touches the horizontal axis to find the total flight time.

Here are some examples of terminal velocities:

1. The linearly determined terminal velocity for an oil drop, such as used in the Millikan oil drop experiment (with approximate diameter 0.0000015 m) is 0.000061 m/s. The quadratic contribution is ten million times smaller and hence negligible, so there is no need to calculate the quadratic term; the air resistance is purely linear.

2. On the other hand, a baseball, as with most larger objects, has a predominately quadratic air resistance of 35 m/s (the linear portion contributes only 1/600 of the quadratic).

3. A drop of mist (with diameter 0.02 mm) has both a linear and a quadratic contribution: the linear terminal velocity is 1.3 m/s and the quadratic terminal velocity is 2.03 m/s. Both linear and quadratic contributions are important [1]. The quadratic term becomes predominant as the water droplet becomes larger, as with most larger objects.

As an object becomes larger, both linear and quadratic terminal velocities change [1].

Reference

[1] John R. Taylor, *Classical Mechanics*, Sausalito, CA: University Science Books, 2005, Chapter 2.

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